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Chapter 6. Applications of
Newton’s Laws
Frictional Forces
 Two types:
- static – applies to stationary objects
- kinetic – applies to sliding (moving)
objects
 Like FN, the Frictional Force is a contact
force, but acts parallel to the interface of two
objects
y
FA
FN
Apply Newton’s 2nd Law
f
x
mg
F
y
 FN  mg  may  0
FN  mg
F
x
 FA  f  max
 If applied force is small, book does not move
(static), ax=0, then f=fs
FA  f S
 Increase applied force, book still does not move
 Increase FA more, now book moves, ax0
FA  f S  max  FA  f S  max  f S
 There is some maximum static frictional force,
fsmax. Once the applied force exceeds it, the book
moves
Magnitudes
f
max
S
  S FN
not vectors
 s is the coefficient of static friction, it is a
dimensionless number, different for each
surface-object pair (wood-wood, wood-metal);
also depends on surface preparation
 s does not depend on the mass or surface
area of the object
 Has value: 0 < s < 1.5
 If no applied vertical force
f
max
S
  S mg
 Push down on
book
y
FN
FA
 Apply Newton’s
2nd Law
F
y
FP
f
x
mg
 FN  mg  FP  ma y  0
FN  mg  FP
f
max
S
  S FN   S (mg  FP )
 What is FA needed just to start book moving?
F
x
 FA  f S  0  FA  f
max
S
 S (mg  FP )
When an Object is Moving?
 fsmax is exceeded so the object can move, but
friction force is still being applied.
 However, less force is needed to keep an object
moving (against friction) than to get it started
 We define kinetic friction
f k   k FN
 k is the coefficient of kinetic friction, similar to
S but always less than S
 Now, let’s consider incline plane problem, but
with friction
 Book is at rest
 m = 1.00 kg,  = 10.0°, s = 0.200
y
fs
F
y
FBD
fs
FN
mg

y
FN

mg
x
F
 FN  mg cos  0
x
FN  mg cos

x
 mg sin   f S  0
f S  mg sin 
o
m

(
1
.
00
kg
)(
9
.
80
)(sin
10
.
0
)
 (1.0kg)(9.80 )(cos 10.0 )
s2
m
s2
 9.65 N
o
 1.70 N
 Book can move (slide) if fs>fsmax
 What is fsmax?
f
max
S
  S FN  (0.200)(9.65 N)
 1.93 N  f S
Book does not move.
 What angle is needed to cause book to slide?
fS  f
max
S
mg sin    S FN
mg sin    S mg cos
tan    S
  tan 1 (  S )
 11.3
o
 As  is increases, FN
decreases, therefore
fsmax decreases
 Once book is moving, we need to use the
kinetic coefficient of friction
 Lets take,  = 15.0° and k = 0.150 < s
F
x
 mg sin   f k  max
 mg sin    k mg cos  max
or a x  g (sin    k cos )
 (9.80 )(sin 15.0  0.150 cos15.0 )
m
s2
 1.12 sm2
o
o
The Tension Force
rope
Crate, m
FBD of crate
T
T
mg
frictionless
Frictionless
pulley
 Assume rope is massless
and taut
FBD of rope
T
T
-T
F
x
 T T  0
m
Crate at rest
F
T
mg
 T  mg  0
T  mg
y
 Like the normal force, the friction and tension
forces are all manifestations of the electromagnetic
force
 They all are the result of attractive (and
repulsive) forces of atoms and molecules within an
object (normal and tension) or at the interface of
two objects
Applications of Newton’s 2nd Law
 Equilibrium – an object which has zero
acceleration, can be at rest or moving with
constant velocity

F  0  F
x
 0,  Fy  0
 Example: book at rest on an incline with friction
 Non-equilibrium – the acceleration of the
object(s) is non-zero


 F  ma   Fx  max ,  Fy  may
Example Problem
Three objects are connected by strings that pass
over massless and frictionless pulleys. The objects
move and the coefficient of kinetic friction between
the middle object and the surface of the table is
0.100 (the other two being suspended by strings).
(a) What is the acceleration of the three objects? (b)
What is the tension in each of the two strings?
 Given: m1 = 10.0 kg,
m2=80.0 kg, m3=25.0
kg, k=0.100
m1
m2
m3
 Find: a1, a2, a3, T1,
and T2
Solution: 1) Draw free-body diagrams
T1
m1
T2
m3
FN
T1
y
T2
fk
m1g
m3 g
m2g
2) Apply Newton’s 2nd Law to each object
x
F
y1
 m1a y1
F
y3
 m3 a y 3
T1  m1 g  m1a y1
T2  m3 g  m3 a y 3
T1  m1 (a y1  g )
T2  m3 (a y 3  g )
F
x2
 m2 ax 2
F
y2
0
T2  T1  f k  m2 ax 2 FN  m2 g  0
Also, ax1=0, ax3=0, ay2=0
FN  m2 g
f k   k FN   k m2 g
T2  T1   k m2 g  m2 a x 2
T1  m1 (a y1  g )
T2  m3 (a y 3  g )
 Three equations, but
five unknowns:
ay1,ay3,ax2,T1,and T2
 But, ay1 = ax2 = -ay3 = a
 Substitute 2nd and 3rd equations into the 1st
equation
m3 (a  g )  m1 (a  g )   k m2 g  m2 a
 m3 a  m3 g  m1a
 m1 g   k m2 g  m2 a  0
a(m3  m1  m2 )
 g (m3  m1   k m2 )  0
a(m1  m2  m3 )  g (m3  m1   k m2 )
(m3  m1   k m2 )
ag
(m1  m2  m3 )
 25.0  10.0  80.0(0.100) 
 9.80

 10.0  80.0  25.0

 0.60 sm2  a y1  a x 2  a y 3
T1  m1 (a  g )  10.0(0.60  9.80)
 104 N
T2  m3 ( g  a)  25.0(9.80  0.60)
 230 N
Example Problem
A skier is pulled up a slope at a constant velocity by
a tow bar. The slope is inclined at 25.0° with respect
to the horizontal. The force applied to the skier by
the tow bar is parallel to the slope. The skier’s mass
is 55.0 kg , and the coefficient of kinetic friction
between the skis and the snow is 0.120. Find the
magnitude of the force that the tow bar exerts on
the skier.
 Given: m = 55.0 kg, k = 0.120,  = 25.0°
 Infer: since velocity is constant, ax=0; also
ay=0 since skier remains on slope
  equilibrium
F  0  F
x
 0,  Fy  0
Draw FBD, apply Newton’s 2nd Law
Fp
Fp
FN
FN
fk
mg

fk
x

mg

x
F
y
0
FN  mg cos   0  Fx  0
FN  mg cos  f k  FP  mg sin   0
FP  f k  mg sin 
  k FN  mg sin 
  k mg cos   mg sin 
 mg (  k cos   sin  )
 286 N